Electronic Journal of Qualitative Theory of Differential Equations 2011, No. 56, 1-14;http://www.math.u-szeged.hu/ejqtde/
Multiple solutions of nonlocal boundary value problems for fractional differential equations on
the half-line ∗
Xiping Liu, Mei Jia
†College of Science, University of Shanghai for Science and Technology, Shanghai, 200093, China
Abstract: In this paper, we study the existence of multiple solutions of nonlocal boundary value problems for fractional differential equations with integral boundary conditions on the half-line.
Applying the fixed point theory and the upper and lower solutions method, some new results on the existence of at least three nonnegative solutions are obtained. An example is presented to illustrate the application of our main results.
Keywords: Fractional differential equations; Caputo derivative; Integral boundary condition;
Lower and upper solutions; Half-line.
MSC:34B15, 26A33.
1 Introduction
In this paper, we consider the following nonlocal boundary value problem for fractional differ- ential equations with integral boundary condition on the half-line
CDα(p(t)u′(t)) +q(t)f(t, u(t)) = 0, t >0, p(0)u′(0) = 0,
t→lim+∞u(t) =R+∞
0 g(t)u(t)dt,
(1.1)
where CDα is the standard Caputo derivative, 0< α < 1 is a constant, f, g, pand q are given functions.
Boundary value problems (BVPs) of differential equation have received much attention in recent years due to their broad applications in applied mathematics and physics. There are many papers
∗Supported by Natural Science Foundation of China (No. 11071164) and the Innovation Program of Shanghai Municipal Education Commission (No. 10ZZ93).
†Corresponding author E-mail address: [email protected] (M. Jia).
concerning the existence of solutions, positive solutions or multiple solutions of two point BVPs, three point BVPs,m-point even nonlocal boundary conditions such as integral boundary conditions about the integer order differential equation. For details we can refer to [6, 10, 13, 16–21, 24, 26].
Boundary value problems on the half-line have been applied in unsteady flow of gas through a semi- infinite porous medium, the theory of drain flows, etc. In the paper [1], Agarwal and O’Regan gave infinite interval problems modeling phenomena which arise in the theory of plasma and electrical potential theory. In [6, 10, 11, 13, 19, 26], authors studied two-point or multipoint boundary value problems on the half-line by using different method. The papers [20, 21] studied the existence of positive solutions for second-order boundary value problems of differential equations system with integral boundary condition on the half-line.
It is well known that fractional order differential equations have been proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering, and they also have been of great interests, see [9, 15]. Recently, there are some papers which deal with the existence of the solutions of the boundary values problems for fractional differential equations on finite intervals. For details, see [2, 4, 5, 7, 8, 12, 14, 23, 25, 27, 28] and the references therein.
In [9] and [15], the basic theories for the fractional calculus and the fractional differential equations were discussed. In [5], Benchohra, Hamania and Ntouyas investigated the existence and uniqueness of solutions for problem:
CDαy(t) =f(t, y(t)), t∈[0, T], 1< α≤2, y(0) =g(y), y(T) =yT.
By using Schauder’s fixed point theorem combined with the diagonalization method, Arara and co-authors (see [4]) studied the existence of solutions for boundary value problems for fractional order differential equation of the form
CDαy(t) +f(t, y(t)) = 0, t∈[0,∞), 1< α≤2, y(0) =y0, yis bounded on [0,∞).
In [2], Ahmad and Nieto stuided some existence results for a boundary value problem involving a nonlinear integrodifferential equation of fractional order 1 < q ≤ 2 with integral boundary conditions by using contraction mapping principle and Krasnoselski´ı’s fixed point theorem.
However, researches for the multiple solutions of the fractional differential equations with non- local boundary condition on infinite intervals are few. In this paper, we aim to discuss the multiple solutions for fractional differential equations with integral boundary condition on the half-line.
Applying the well-known Amann theorem and the upper and lower solutions method, we obtain a new result on the existence of at least three distinct nonnegative solutions under some conditions.
An example is presented to illustrate the application of our main result.
2 Preliminaries
In this section, we introduce preliminary facts which are used throughout this paper. We denote that R= (−∞,+∞) andR+= [0,+∞).
Definition 2.1(See [9, 15]) Letα > 0. The fractional (arbitrary) order integral of the function y:R+ →Rof order αis defined by
Iαy(t) = 1 Γ(α)
Z t 0
(t−s)α−1y(s)ds.
provided the integral exists, where Γ is the Gamma function.
Definition 2.2(See [9, 15]) The Caputo fractional order derivative of the functiony of orderαis defined by
CDαy(t) = 1 Γ(n−α)
Z t 0
y(n)(s) (t−s)α+1−nds,
provided the right side is pointwise defined on (0,+∞), where n = [α] + 1 and [α] denotes the integer part ofα.
Throughout the paper, we suppose that the following hypotheses are satisfied:
(H1)g∈L1(R+),g(t)≥0,t∈R+, and 0≤R+∞
0 g(t)dt:=||g||1<1.
(H2)p(t)>0 for allt∈R+,R+∞ 0
1
p(r)drexists and the functionk(s) =R+∞ s
(r−s)α−1
p(r) dr <+∞ is continuous onR+.
It is obvious that 0<1− ||g||1≤1 if (H1) holds. From (H2), we can get that k(s)≥0 and
s→lim+∞k(s) = 0. Sok(s) is bounded, which implies that there exists a constantK0>0 such that 0≤k(s)≤K0= sup
s∈R+
k(s), fors∈R+. We define that
K(t, s) =
R+∞
t
(r−s)α−1
p(r) dr, 0≤s < t, R+∞
s
(r−s)α−1
p(r) dr, 0≤t≤s,
(2.1) and
G(t, s) = 1
(1− ||g||1)Γ(α)
(1− ||g||1)K(t, s) + Z +∞
0
g(r)K(r, s)dr
. (2.2)
By (H1), (H2), (2.1) and (2.2), we can easily get thatKandGsatisfy the following lemma.
Lemma 2.1 Suppose (H1) and (H2) hold. Then
(1) K(t, s)is well defined and continuous, for all(t, s)∈R+×R+; (2) 0≤K(t, s)≤K(s, s) =k(s)≤K0, for all(t, s)∈R+×R+;
(3) G(t, s) is well defined, continuous, and 0≤G(t, s)≤G(s, s)≤ (1−||g||11)Γ(α)K0, for all(t, s)∈ R+×R+;
(4) For anys∈R+, lim
t→+∞K(t, s) = 0, and denote G∞(s) := lim
t→+∞G(t, s) = 1 (1− ||g||1)Γ(α)
Z +∞ 0
g(r)K(r, s)dr, then G∞(s)is continuous, and
G∞(s)≤ 1
(1− ||g||1)Γ(α)K0.
Lemma 2.2 Suppose that (H1) and (H2) hold, and h∈ L1(R+). Then the following boundary value problem
CDα(p(t)u′(t)) +h(t) = 0, t >0, p(0)u′(0) = 0,
u(∞) =R+∞
0 g(t)u(t)dt,
(2.3)
has a unique solution
u(t) = Z +∞
0
G(t, s)h(s)ds, whereu(∞) := lim
t→+∞u(t).
Proof. By (2.3), we have
p(t)u′(t) =p(0)u′(0)−Iαh(t) =− 1 Γ(α)
Z t 0
(t−s)α−1h(s)ds.
We getu′(t) =−Γ(α)p(t)1 Rt
0(t−s)α−1h(s)ds, and u(t) =u(0)−
Z t 0
1 Γ(α)p(r)
Z r 0
(r−s)α−1h(s)ds
dr
=u(0)− 1 Γ(α)
Z t 0
dr Z r
0
(r−s)α−1h(s)
p(r) ds
=u(0)− 1 Γ(α)
Z t 0
ds Z t
s
(r−s)α−1h(s)
p(r) dr. (2.4)
So
u(∞) =u(0)− 1 Γ(α)
Z +∞ 0
ds Z +∞
s
(r−s)α−1h(s)
p(r) dr=
Z +∞ 0
g(t)u(t)dt.
Then
u(0) = Z +∞
0
g(r)u(r)dr+ 1 Γ(α)
Z +∞ 0
ds Z +∞
s
(r−s)α−1h(s)
p(r) dr. (2.5)
Substituting (2.5) into (2.4), we have u(t) =
Z +∞ 0
g(r)u(r)dr+ 1 Γ(α)
Z +∞ 0
ds Z +∞
s
(r−s)α−1h(s)
p(r) dr− 1
Γ(α) Z t
0
ds Z t
s
(r−s)α−1h(s)
p(r) dr
= Z +∞
0
g(r)u(r)dr+ 1 Γ(α)
Z t 0
ds Z +∞
t
(r−s)α−1h(s)
p(r) dr+ 1
Γ(α) Z +∞
t
ds Z +∞
s
(r−s)α−1h(s)
p(r) dr
= Z +∞
0
g(r)u(r)dr+ 1 Γ(α)
Z t 0
Z +∞ t
(r−s)α−1 p(r) dr
h(s)ds+ 1 Γ(α)
Z +∞ t
Z +∞ s
(r−s)α−1 p(r) dr
h(s)ds.
By (2.1), we get that
u(t) = Z +∞
0
g(r)u(r)dr+ 1 Γ(α)
Z +∞ 0
K(t, s)h(s)ds. (2.6)
So
g(t)u(t) =g(t) Z +∞
0
g(r)u(r)dr+ g(t) Γ(α)
Z +∞ 0
K(t, s)h(s)ds.
Z +∞ 0
g(t)u(t)dt= Z +∞
0
g(t)dt· Z +∞
0
g(r)u(r)dr+ 1 Γ(α)
Z +∞ 0
g(r) Z +∞
0
K(t, s)h(s)ds dr
= Z +∞
0
g(t)dt· Z +∞
0
g(r)u(r)dr+ 1 Γ(α)
Z +∞ 0
ds Z +∞
0
g(r)K(r, s)h(s)dr
= Z +∞
0
g(t)dt· Z +∞
0
g(r)u(r)dr+ 1 Γ(α)
Z +∞ 0
Z +∞ 0
K(r, s)g(r)dr h(s)ds.
Noting that 0<1− ||g||1= 1−R+∞
0 g(t)dt≤1, we have Z +∞
0
g(r)u(r)dr= 1
(1− ||g||1)Γ(α) Z +∞
0
Z +∞ 0
K(r, s)g(r)dr
h(s)ds. (2.7) Substituting (2.7) into (2.6), we have
u(t) = 1
(1− ||g||1)Γ(α) Z +∞
0
Z +∞ 0
K(r, s)g(r)dr
h(s)ds+ 1 Γ(α)
Z +∞ 0
K(t, s)h(s)ds
= 1
(1− ||g||1)Γ(α) Z +∞
0
Z +∞ 0
K(r, s)g(r)dr+ (1− ||g||1)K(t, s) h(s)ds
= Z +∞
0
G(t, s)h(s)ds, where
G(t, s) = 1
(1− ||g||1)Γ(α) Z +∞
0
K(r, s)g(r)dr+ (1− ||g||1)K(t, s) .
Lemma 2.3 Suppose (H1) and (H2) hold, if u=u(t)satisfies
CDα(p(t)u′(t))≤0, t∈(0,+∞), p(0)u′(0)≤0,
u(∞)−R+∞
0 g(r)u(r)dr≥0.
(2.8)
Thenu(t)≥0 for t∈R+.
Proof. LetCDα(p(t)u′(t)) =−h(t)≤0,p(0)u′(0) =k0≤0 andu(∞)−R+∞
0 g(r)u(r)dr=k1≥ 0. We consider the following boundary value problem
CDα(p(t)u′(t)) =−h(t), t∈(0,+∞), p(0)u′(0) =k0,
u(∞)−R+∞
0 g(r)u(r)dr=k1.
(2.9)
Similar to the proof of Lemma 2.2, we can obtain that the BVP (2.9) has a unique solution.
u(t) = k1
(1− ||g||1)− k0
(1− ||g||1)
Z +∞ 0
ds Z +∞
s
g(s)
p(r)dr+ (1− ||g||1) Z +∞
t
dr p(r)
+
Z +∞ 0
G(t, s)h(s)ds.
(2.10) Sincek0≤0,k1≥0 andh(t)≥0 for t >0, it is easy to show
u(t)≥0, for t∈R+
from (2.10), (H1) and (H2).
LetE be a Banach space,P ⊂Ebe a cone inE. A coneP is called solid if it contains interior points, i.e., ˚P 6= Ø. Every coneP inE defines a partial ordering inEgiven byxy iffy−x∈P. Ifxy andx6=y, we writexy; if a coneP is solid andy−x∈P˚, we writex≪y. A coneP is said to be normal if there exists a constantN >0 such that 0xy implies||x|| ≤N||y||. If P is normal, then every order interval [x, y] ={z∈E|xzy} is bounded.
The following Lemma 2.4 is the well-known Amann three-solution theorem (see [3, 22]), which will be used in the later proof of our main results about the multiple solutions of the boundary value problem.
Lemma 2.4 LetE be a Banach space, andP be a normal solid cone. Suppose that there existα1, β1,α2,β2∈E with
α1β1α2β2,
andT : [α1, β2]−→E is a completely continuous strongly increasing operator such that α1T α1, T β1β1, α2T α2, T β2β2.
Then the operator T has at least three fixed points x1, x2, x3 such that α1x1≪β1, α2≪x2β2, α2x3β1.
3 Multiple solutions of the boundary value problem
Definition 3.1. u=u(t) is called an upper (lower) solution of boundary value problem (1.1), if
it satisfies
CDα(p(t)u′(t)) +q(t)f(t, u(t))≤0 (≥0), t >0, p(0)u′(0)≤0 (≥0),
u(∞)−R+∞
0 g(t)u(t)dt≥0 (≤0).
In order to obtain the results, we suppose the following conditions hold:
(H3)q∈L1(R+),q(t) is nonnegative onR+ andq >0 a.e. onR+.
(H4)f :R+×R+−→R+ is a Carath´eodory function, that is to say,f(·, u) is measurable for anyu∈R+ andf(t,·) is continuous for almost everyt∈R+. f(t, u) is bounded fort∈R+ when uis bounded, and
f(t, u1)< f(t, u2) with u1< u2∈R+, for almost every t∈R+. LetE={u∈C(R+)| lim
t→+∞u(t) exists}be endowed with the norm kuk:= sup
t∈R+|u(t)|, then E is a Banach space. We define the coneP ⊂E by
P:={u∈E|u(t)≥0, t∈R+}.
Obviously, P is a normal solid cone in E, and uv ∈ E if and only ifu(t)≤ v(t) for t ∈R+. uv ∈E if and only if u(t)≤v(t) and u(t)6≡v(t), which implies that there exists an interval [a0, b0]⊂R+ such thatu(t)< v(t) fort∈[a0, b0].
Lemma 3.1 (See [11]) LetE be defined as before and D⊂E. ThenD is relatively compact inE if the following conditions hold:
(a)D is uniformly bounded in E;
(b)the functions fromD are equicontinuous on any compact interval of[0,+∞);
(c) the functions from D are equiconvergent, that is, for any given ε > 0, there exists a R0 = R(ε)>0 such that |u(t)−u(+∞)|< ε, for anyt > R0,u∈D.
Now, we define an operatorT : P −→E by (T u)(t) =
Z +∞ 0
G(t, s)q(s)f(s, u(s))ds.
Lemma 3.2 Suppose that (H1)–(H4) hold. Then the operatorT : P −→P, and it is completely continuous.
Proof. First of all, let us show the operatorT is well defined, andT : P−→P.
For any fixedu∈P, it implies thatuis bounded, by Lemma 2.1, (H3) and (H4), and we can get (T u)(t)≥0 for t∈R+. And there exists a constantfM0 >0 such that 0 ≤f(t, u(t))≤fM0
for anyt∈R+. Then 0≤(T u)(t) =
Z +∞ 0
G(t, s)q(s)f(s, u(s))ds≤ K0fM0
(1− ||g||1)Γ(α) Z +∞
0
q(s)ds <+∞. And
t→lim+∞(T u)(t) = lim
t→+∞
Z +∞ 0
G(t, s)q(s)f(s, u(s))ds= 1 (1− ||g||1)Γ(α)
Z +∞ 0
G∞(s)q(s)f(s, u(s))ds <+∞. Thus,T : P−→P is well defined.
Secondly, we show thatT is continuous.
Let{un} ⊂P, u∈P, and un →u0as n→ ∞. So, there exists a constantfM1 >0, such that 0≤f(t, un(t)), f(t, u0(t))≤fM1 for anyt∈R+. By (H3), (H4) and Lemma 2.1, we can see
||T un−T u0||= sup
t∈R+|(T un)(t)−(T u0)(t)| ≤ Z +∞
0
G(s, s)q(s)|f(s, un(s))−f(s, u0(s))|ds,
G(s, s)q(s)|f(s, un(s))−f(s, u0(s))| ≤ 2K0fM1
(1− ||g||1)Γ(α)q(s) and
nlim→∞ f(s, un(s))−f(s, u0(s)
= 0, a.e. s∈R+. According to the Lebesgue’s dominated convergence theorem, we can show
nlim→∞
Z +∞ 0
G(s, s)q(s)|f(s, un(s))−f(s, u0(s))|ds= 0.
Therefore, the operatorT is continuous.
Finally, we will prove that the operatorT maps bounded sets into relatively compact sets.
For the bounded set Ω⊂P, there exists a constantM2>0, such that||u|| ≤M2for anyu∈Ω.
Thus there exists a constant fM2 >0, such that 0≤f(t, u(t))≤fM2 for anyt∈R+. And 0≤(T u)(t) =
Z +∞ 0
G(t, s)q(s)f(s, u(s))ds≤ K0fM2
(1− ||g||1)Γ(α) Z +∞
0
q(s)ds <+∞. Thus, the setT(Ω) is uniformly bounded.
For any [a, b]⊂[0,+∞) and anyt1, t2∈[a, b], by Lemma 2.1, we haveG(t1, s)−G(t2, s)→0, as t1→t2for anys∈R+. And
0≤ |G(t1, s)−G(t2, s)|q(s)f(s, u(s))≤2G(s, s)q(s)fM2 ≤ 2K0fM2
(1− ||g||1)Γ(α)q(s).
Then
|(T u)(t1)−(T u)(t2)|=| Z +∞
0
G(t1, s)q(s)f(s, u(s))ds− Z +∞
0
G(t2, s)q(s)f(s, u(s))ds|
≤ Z +∞
0 |G(t1, s)−G(t2, s)|q(s)f(s, u(s))ds
≤ Z +∞
0 |G(t1, s)−G(t2, s)|q(s)fM2ds
→0, as t1→t2. (3.1)
That is, T ufrom T(Ω) is equicontinuous on any compact interval of [0,+∞).
By Lemma 2.1, we have
|(T u)(t)−(T u)(∞)|=| Z +∞
0
G(t, s)−G∞(s)
q(s)f(s, u(s))ds|
≤ Z +∞
0 |G(t, s)−G∞(s)|q(s)fM2ds
→0, as t→+∞.
ThenT ufromT(Ω) is equiconvergent.
Using Lemma 3.1, we can obtain that the set T(Ω) is a relatively compact set. Hence, the operatorT maps bounded sets into relatively compact sets.
Therefore, we can get that the operatorT is completely continuous.
Theorem 3.3 Suppose that (H1)–(H4) hold, and there exist two lower solutions x1, x2 and two upper solutionsy1, y2 of boundary value problem(1.1)such thatx2, y1 are not the solutions of the boundary value problem(1.1)with
x1y1x2y2.
Then the boundary value problem(1.1)has at least three distinct nonnegative solutionsu1,u2,u3
which satisfy that fort∈R+
x1(t)≤u1(t)< y1(t), x2(t)< u2(t)≤y2(t), x2(t)u3(t)y1(t).
Proof. It is obvious that the boundary value problem (1.1) has nonnegative solutions if and only if the operatorT has fixed points onP.
It follows from Lemma 3.2 thatT : [x1, y2]→P is completely continuous.
Let us prove thatT is a strongly increasing operator.
For anyw1, w2∈P, with w1 w2, that is to say thatw1(t)≤w2(t) for allt∈R+, and there exists [a0, b0]⊂R+ such thatw1(t)< w2(t) for anyt∈[a0, b0].
Hence, for anyt∈R+, using the conditions (H3) and (H4) we have (T w2)(t)−(T w1)(t) =
Z +∞ 0
G(t, s)q(s) f(s, w2(s))−f(s, w1(s)) ds
≥ Z b0
a0
G(t, s)q(s) f(s, w2(s))−f(s, w1(s)) ds
>0.
We can get that
0<(T w2)(t)−(T w1)(t)∈P .˚ Hence, we conclude that T is a strongly increasing operator.
Let us now prove thatx1T x1. We denotex=T x1−x1.
Noting thatx1is the lower solution of boundary value problems (1.1) and applying the definition
of the operatorT, we have
CDα(p(t)x′(t)) =CDα
p(t) (T x1)′(t)−x′1(t)
=CDα(p(t)(T x1)′(t))−CDα(p(t)x′1(t))
=−q(t)f(t, x1(t))−CDα(p(t)x′1(t))
≤0,
x′(0) = (T x1)′(0)−x′1(0)≤0, and
x(∞)− Z +∞
0
g(t)x(t)dt= (T x1)(∞)−x1(∞)− Z +∞
0
g(t) (T x1)(t)−x1(t) dt≥0.
It follows from Lemma 2.3 that
x(t) = (T x1)(t)−x1(t)≥0, for t∈R+. Then
x1T x1. Similarly, we can get that
x2T x2.
Sincex2 is an lower solution of (1.1) and not a solution of (1.1), we have (T x2)6=x2. Thus x2T x2.
Using the same method, we can also get that
T y1y1, T y2y2.
Using the Lemma 2.4, we obtain T has at least three fixed points u1, u2, u3 ∈ [x1, y2].
Moreover,u1, u2, u3∈P and
x1u1≪y1, x2≪u2y2, x2u3y1.
Hence, the boundary value problem (1.1) has at least three distinct nonnegative solutionsu1, u2, u3∈ [x1, y2] and we see fort∈R+
x1(t)≤u1(t)< y1(t), x2(t)< u2(t)≤y2(t), x2(t)u3(t)y1(t).
4 Illustration
To illustrate our main results, we present an example.
Example 4.1. Consider the following integral boundary value problem
CD12(p(t)u′(t)) +q(t)f(t, u) = 0, t >0, p(0)u′(0) = 0,
u(∞) =R+∞
0 e−2tu(t)dt,
(4.1)
where
p(t) =et, f(t, u) = 200
√π
u2, 0≤u <1,
√u , u≥1.
We takeα= 12,q(t) =te−t, g(t) =e−2t. It is easy to show that g∈L1(R+), 0≤ ||g||1=
Z +∞ 0
g(t)dt= 1
2 <1, and 1− ||g||1= 1− Z +∞
0
g(t)dt=1 2, then (H1) holds.
p(t) =et>0, k(s) = Z +∞
s
(r−s)α−1 p(r) dr=√
πe−s, s∈R+, that is (H2) holds.
q(t) =te−t, Z +∞
0
q(t)dt= Z +∞
0
te−tdt= 1<+∞, it implies that (H3) holds.
We can easily verify the condition (H4) holds.
In view of Lemma 2.2, for anyh∈ L1(R+), u(t) =R+∞
0 G(t, s)h(s)ds satisfies the boundary conditions of (4.1).
Now, leth(t) = 200t√πe−t. It is obvioush∈L1(R+).
Fort∈R+, we take
x1(t) = 0, x2(t) = 1 242
Z +∞ 0
G(t, s)h(s)ds, and
y1(t) = 1 532
Z +∞ 0
G(t, s)h(s)ds, y2(t) = 53 Z +∞
0
G(t, s)h(s)ds.
Thenx1,x2,y1,y2∈P.
It is easy to see 0 =x1(t)< y1(t)< x2(t)< y2(t) fort∈R+, that isx1y1x2y2. Moreover, we have
p(0)x′i(0) = 0, p(0)yi′(0) = 0, xi(∞) = Z +∞
0
e−2txi(t)dt, yi(∞) = Z +∞
0
e−2tyi(t)dt, i= 1,2.
Through calculation, we can get that 24<R+∞
0 G(t, s)h(s)ds <53, and we can easily verify
CD12(p(t)x′1(t)) +q(t)f(t, x1(t)) = 0,
CD12(p(t)x′2(t)) +q(t)f(t, x2(t))>CD12(p(t)x′2(t)) + 1
242h(t) = 0,
CD12(p(t)y′1(t)) +q(t)f(t, y1(t))<CD12(p(t)y′1(t)) + 1
532h(t) = 0 and
CD12(p(t)y2′(t)) +q(t)f(t, y2(t))≤CD12(p(t)y′2(t)) + 53h(t) = 0.
Therefore,x1(t), x2(t) are lower solutions of BVP (4.1), andy1(t), y2(t) are upper solutions of BVP (4.1).
It follows from Theorem 3.3 that the boundary value problem (4.1) has at least three distinct nonnegative solutionsu1, u2, u3∈[x1, y2]. Moreover, fort∈R+
x1(t)≤u1(t)< y1(t), x2(t)< u2(t)≤y2(t), x2(t)u3(t)y1(t).
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(Received February 20, 2011)
